Presentation on theme: "1 PythagoreanTheorem Statement 2 In a right triangle The sum of the areas of the squares on its sides equals the area of the square on its hypotenuse."— Presentation transcript:
2 In a right triangle The sum of the areas of the squares on its sides equals the area of the square on its hypotenuse.
3 Proofs There are several proofs of Pythagorean theorem. Some of them are rigorous analytical proofs. They are based on the properties of triangles, such as congruence of triangles or similarity of triangles. Some others are based on transformations.
4 Analytical Proof We will now discuss a proof based on the congruence of triangles.
5 Consider a right triangle ABC Δ ABC is right angled at C.
6 It is given that triangle ABC is right angled at C We have to prove that Area sq. ACGF + Area sq. BCHK = Area sq. ABDE.
7 Now, consider ΔABF & ΔAEC They are congruent, because AE = AB (why?) They are sides of the same square ABDE AF = AC, (why?) They are sides of the same square ACGF And also, BAF = BAC + CAF = CAB + BAE = CAE Hence by SAS, Δ ABF = Δ AEC …(i).
8 But, the area of the ΔABF is half the area of the square ACGF. Δ ABF has base AF and the altitude from B on it = CA Its area therefore equals half the area of square on the side AF area Δ ABF =½ area sq. ACGF
9 The area of the ΔAEC equals half the area of the rectangle AELM. On the other hand, ΔAEC has base AE The altitude from C = AM, (where M is the point of intersection of AB with the line CL parallel to AE) Therefore, area of ΔAEC = ½ area rectangle AELM …(iii)
10 The area of the square ACGF = the area of the rectangle AELM We have From (i), Δ ABF = Δ AEC From (ii), area of Δ ABF = ½ area of sq. ACGF And from (iii), area of Δ AEC = ½ area of rect. AELM Thus, from (i), (ii) and (iii), area of sq. ACGF =area of rect. AELM **(a).
11 In the same way, Can you establish that The area of sq. BKHC =area of rect. BDLM…?
12 O.K. -- Let us consider ΔABK & ΔDBC They are congruent, because BD = BA (why?) BC = BK, (why?) ABK = ABC + CBK = CBA + ABD = DBC Hence by SAS, Δ ABK = Δ DBC …(iv)
13 The area of the ΔABK equals half the area of the square BKHC Δ ABK has base BK The altitude from A = BC. Therefore, area of Δ ABK = ½ area of square BKHC …(v).
14 The area of the ΔBDC equals half the area of the rectangle BDLM. On the other hand, Δ BDC has base BD The altitude from C = BM, Therefore, the area of Δ BDC = ½ area of rect. BDLM …(vi).
15 Thus, the area of the square on side BC equals the area of the rectangle BDLM We now have Δ ABK = Δ DBC … (iv) area of ΔABK =half area of sq. BKHC.. (v) And area of ΔDBC =half area of rect. BDLM.. (vi) From (iv), (v) and (vi), area of sq. BKHC =area of rect. BDLM **(b).
16 Combining the results area of sq. ACGF =area of rect. AELM **(a) And also, area of sq. BKHC =area of rect. BDLM **(b).
17 Summing UP area of sq. ACGF +area of sq. BKHC =area of rect. AELM +area of rect. BDLM = area of sq. AEDB. In other words, area of sq. on side AC + area of sq. on side BC. = area of the square on hypotenuse AB.
18 Applications Integers which can form the sides of a right triangle are called Phythagorean Triplets. Like 3,4 and 5. And 5,12 and … ??? Think Calculate.
19 Teaser (an extension activity) If two sides AC and BC measure 3 and 4 units -- But if the included angle is not a right angle -- but an obtuse angle, then AB will be….. More than 5. Again, if the angle is acute, then !!!!.
20 Verification We will now see a demonstration of verification of Pythagorean theorem through transformations.